Question

A person is know to speak the truth 4 times out of 5. He throws a die and reports that it is a ace. The probability that it is actually a ace is

• A. $1/3$
• B. $2/9$
• C. $4/9$
• D. $5/9$

Answer 1

The correct option is C $4/9$

Let $E_1$ denote the event that an ace occurs and $E_2$ the event that it does not occur. Let $A$ denote the event that the person reports that it is an ace.

Then $P\left(E_1\right)=1/6,P\left(E_2\right)=5/6,P\left(A|E_1\right)=4/5$ an $P\left(A|E_2\right)=1/5$.

By Bayes' theorem,
$P\left(E_1|A\right)=\frac{P\left(E_1\right)P\left(A|E_1\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}$
$=\frac{4}{9}$